1. Gases Unit 6

2. Kinetic Molecular Theory  Kinetic energy is the energy an object has due to its motion.  Faster object moves = higher kinetic energy  According to the Kinetic Molecular Theory, ALL particles of matter are in constant motion.  This theory helps explain the behavior of solids, liquids, and gases.

3. Kinetic Energy and Temperature  Increase in kinetic energy results in an increase in temperature.  Temperature (in Kelvin) is directly proportional to the average kinetic energy.  Plot: Maxwell-Boltzmann curves

6. Average speed decreases with increasing molar mass

7. Kinetic Theory and Gases This theory includes the following assumptions: 1. The particles in a gas are considered to be small, hard spheres with an insignificant volume. Relative to solids and liquids, the particles in gases are far apart. The particles are not attracted or repulsed by each other, and they move independently.

8. 2. The motion of the particles in a gas is rapid, constant, and random. The particles travel in a straight-line path until they collide with an object or another particle. The particles change direction only when they rebound from a collision.

9. 3. All collisions between particles in a gas are perfectly elastic. This means that during a collision the kinetic energy is transferred without loss from one particle to another. The total kinetic energy remains constant.

10. Pressure  Gas pressure results from the force exerted by a gas per unit surface area of an object.  Pressure is the result of the billions of collisions between gas particles & an object.  Atmospheric pressure results from the collisions of atoms and molecules in air.  More forceful collisions or more frequent collisions mean higher gas pressure.

11. Units of Pressure P = F/A  N/m 2  1 Pa  1 atm = 760 mm Hg = 101.3 kPa  Other units:  1 atm = 760. mm Hg = 760. torr = 29.92 in Hg = 101,300 Pa = 101.3 kPa = 14.7 psi = 76 cm Hg.

12.  Calculate the number of mm Hg in 1.42 atm  Calculate the number of kPa in 0.93 atm.

13. Factors Affecting Gas Pressure 1. Amount of gas = more gas particles results in more collisions which increases the pressure. 2. Volume = compressing a gas into a container results in more collisions which increases the pressure. 3. Temperature = Gas particles will move more with a temperature increase resulting in more collisions which increases the pressure.

14. Gas laws  The factors affecting gases are expressed and used in several “gas laws”  The gas laws allow us to take situations involving gases and solve them mathematically.  The gas laws we will learn are: Boyle’s Law, Charles’s Law, Gay-Lussac’s Law, Combined Gas Law, Ideal Gas Law, Avogadro’s Law, Dalton’s Law, & Graham’s Law.

15. Boyle’s Law  Robert Boyle proposed this law in 1662  If the temperature is constant, as the pressure of a gas increases, the volume decreases and vice versa.  The law states that for a given mass of gas at constant temperature, the volume of the gas varies inversely with pressure. P 1 x V 1 = P 2 x V 2 P 1 x V 1 = P 2 x V 2

16. Charles’s Law  Jacques Charles in 1787  As the temperature of an enclosed gas increases, the volume increases, if the pressure is constant.  The law states that the volume of a fixed mass of gas is directly proportional to its Kelvin temperature if the pressure is constant. V 1 V 2 V 1 V 2 --- = --- T 1 T 2  K = °C + 273  STP = 0°C 101.3 kPa

17. Gay-Lussac’s Law  Joseph Gay-Lussac in 1802.  As the temperature of an enclosed gas increases, the pressure increases, if the volume is constant.  This law states that the pressure of a gas is directly proportional to the Kelvin temperature if the volume remains constant. P 1 P 2 P 1 P 2 --- = --- T 1 T 2

18. Example 1.00 L of a gas at 1.25 atm is compressed to 473 mL. What is the new pressure of the gas? p 1 V 1 = p 2 V 2 → 1.00 x 1.25 = p 2 x 0.473, so p 2 = 1.00 x 1.25 / 0.473 = 2.64 atm The temperature of 2.5 L of gas is lowered from 25 o C to 5 o C. What is the new volume? V 1 / T 1 = V 2 / T 2 → 2.5 / 298 = V 2 / 278, so V 2 = 278 x 2.5 / 298 = 2.3 L

19. Practice The pressure of 2.50 L of a gas is increased from 0.85 atm to 1.35 atm. What is the new volume of the gas? The pressure of a gas at 275 K is increased from 1.25 atm to 2.25 atm. What is the new temperature?

20. The Combined Gas Law  This is a single expression that combines Boyle’s, Charles’s, and Gay-Lussac’s laws.  This law allows you to do calculations for situations in which only the amount of gas is constant. P 1 x V 1 P 2 x V 2 P 1 x V 1 P 2 x V 2 ----------- = ----------- T 1 T 2 T 1 T 2

21. Practice  The temperature of 4.5 L of a gas at 1.2 atm is decreased from 350 K to 300 K, while the volume increases to 5.5 L. Calculate the new pressure of the gas.

22. Avogadro’s Law  As volume increases, the number of gas molecules increases (constant p and T)  Count number of gas molecules by moles  One mole of any ideal gas occupies 22.4 L at standard conditions - molar volume  Equal volumes of gases contain equal numbers of molecules It doesn’t matter what the gas is! It doesn’t matter what the gas is!

23. Ideal Gas Law  To calculate the number of moles of a contained gas requires an expression with the variable n. PV = nRT R is the ideal gas constant = R is the ideal gas constant = 8.31 (L* kPa) / (K * mol) 8.31 (L* kPa) / (K * mol)  Pressure in kPaVolume in L  Temperature in Kn in moles

24. Example  What volume will 3.15 mol of hydrogen gas occupy at STP?  pV = nRT  101.3 x V = 3.15 x 8.31 x 273  V = 3.15 x 8.31 x 273 / 101.3 = 70.5 L

25. Practice  What pressure is exerted by 1.24 moles of a gas stored in a 15.0 L container at 22 o C?  If you wish to collect 10.0 L of nitrogen gas at 15 o C and 0.945 atm, what mass of gas will you need to have?

27. Stoichiometry  At STP, 1 mole of any gas occupies 22.4 L  Practice:  5.6 L N 2 reacts with excess H 2. How many L of NH 3 will form?

28.  To calculate the volume of a gas at non STP conditions, use the combined gas law with STP and 22.4 L as one of the conditions.  Example: How many L of CO 2 is produced when 25 g of CH 4 is burned at 1.5 atm and 25 o C?

29. Density of gases  pV = nRT and n=mass(m)/molar mass(M), so pV = (m/M)RT  d = m/V, so d = pM/RT  Density is directly proportional to molar mass  M = dRT/p

30. Dalton’s Law  In a mixture of gases, the total pressure is the sum of the partial pressures of the gases.  This law states that at constant volume and temperature, the total pressure exerted by a mixture of gases is equal to the sum of the partial pressures of the component gases. P total = P 1 + P 2 + P 3 + …….

31. Dalton’s law  Mixture of gases: for each gas pV=nRT  With constant V, T the partial pressure of each gas is directly proportional to n.  P total = p a +p b +p c + …. = n a (RT/V) + n b (RT/V) + n c (RT/V) + … = (n a +n b +n c +…)(RT/V) = n total (RT/V)

32. Mole fraction For mixture: mole fraction, X = number of moles of substance / total number of moles X a = n a /n total Thus p a = X a p total

33. Graham’s Law  Thomas Graham during the 1840’s  Gases of lower molar mass diffuse and effuse faster than gases of higher molar mass. 1.Diffusion – particles move from an area of higher particle concentration to an area of lower particle concentration. Diffusion 2. Effusion – same as diffusion except the gas particles escape through small holes (pores) in a container. Effusion

34. Rate diffusion effusion  The rate of effusion/diffusion is inversely proportional to the square root of the molar mass: rate = √(1/molar mass)  Rate gas 1/rate gas 2 = (√M gas 2 (/√(M gas 1 )

35. Real vs. Ideal gases  The Kinetic Molecular Theory explains the behavior of gases using “ideal” gases as a model. Real gases, however, do not always follow this.

36. Ideal gases  Negligible volume  No attractive or repulsive forces  Move randomly in straight lines  Perfectly elastic collisions  (H 2 and He always behave as predicted)

37. Ideal Gases  The ideal gas law holds closely at high temperatures (above 0 o C) and low pressure (below 1 atm).  Under these conditions particles are far enough apart that they display ideal behavior (no interaction and no volume).

38. Real Gases  Attractive forces can not always be disregarded. Water vapor molecules attract one another to form rain or snow in the atmosphere. Water vapor molecules attract one another to form rain or snow in the atmosphere. Volume of gases is not negligible under high pressure.

39. Real gases  Under high pressure particles are pushed close enough together to allow particle interactions, lowering the pressure to some extent (kinetic energy becomes less, so collisions happen with less force).  At lower temperatures the results are similar to when there is high pressure: Less kinetic energy and particle interaction.

40. Kinetic energy and speed For particle: Kinetic energy (KE) = ½ mv 2 For mole: KE = ½ Nmv 2 N = 6.02 x 10 23 Molar mass (M) = N m, so KE = ½ Mv 2 Experiments showed that KE = 3/2 RT with R = 8.31J/K.mol

41. Combining the 2 KE equations: √v 2 = √(3RT/M) √v 2 is called the root-mean-square speed or rms speed This rms speed is related to T and M